Integrand size = 29, antiderivative size = 68 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {a^2 \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}-\frac {3 a^5}{4 d \left (a^3-a^3 \sin (c+d x)\right )} \] Output:
1/4*a^2*arctanh(sin(d*x+c))/d+1/4*a^4/d/(a-a*sin(d*x+c))^2-3/4*a^5/d/(a^3- a^3*sin(d*x+c))
Time = 0.13 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.57 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {a^2 \left (\text {arctanh}(\sin (c+d x))+\frac {-2+3 \sin (c+d x)}{(-1+\sin (c+d x))^2}\right )}{4 d} \] Input:
Integrate[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^2*Tan[c + d*x]^2,x]
Output:
(a^2*(ArcTanh[Sin[c + d*x]] + (-2 + 3*Sin[c + d*x])/(-1 + Sin[c + d*x])^2) )/(4*d)
Time = 0.32 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.84, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3315, 27, 99, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \tan ^2(c+d x) \sec ^3(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^2 (a \sin (c+d x)+a)^2}{\cos (c+d x)^5}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {a^5 \int \frac {\sin ^2(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^3 \int \frac {a^2 \sin ^2(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {a^3 \int \left (\frac {a}{2 (a-a \sin (c+d x))^3}+\frac {1}{4 \left (a^2-a^2 \sin ^2(c+d x)\right )}-\frac {3}{4 (a-a \sin (c+d x))^2}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 \left (\frac {\text {arctanh}(\sin (c+d x))}{4 a}+\frac {a}{4 (a-a \sin (c+d x))^2}-\frac {3}{4 (a-a \sin (c+d x))}\right )}{d}\) |
Input:
Int[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^2*Tan[c + d*x]^2,x]
Output:
(a^3*(ArcTanh[Sin[c + d*x]]/(4*a) + a/(4*(a - a*Sin[c + d*x])^2) - 3/(4*(a - a*Sin[c + d*x]))))/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Result contains complex when optimal does not.
Time = 0.50 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.50
method | result | size |
risch | \(\frac {i a^{2} \left (-4 i {\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}\) | \(102\) |
derivativedivides | \(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2} \sin \left (d x +c \right )^{4}}{2 \cos \left (d x +c \right )^{4}}+a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(167\) |
default | \(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2} \sin \left (d x +c \right )^{4}}{2 \cos \left (d x +c \right )^{4}}+a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(167\) |
Input:
int(sec(d*x+c)^3*(a+a*sin(d*x+c))^2*tan(d*x+c)^2,x,method=_RETURNVERBOSE)
Output:
1/2*I*a^2/d/(exp(I*(d*x+c))-I)^4*(-4*I*exp(2*I*(d*x+c))+3*exp(3*I*(d*x+c)) -3*exp(I*(d*x+c)))+1/4*a^2/d*ln(exp(I*(d*x+c))+I)-1/4*a^2/d*ln(exp(I*(d*x+ c))-I)
Time = 0.13 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.84 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {6 \, a^{2} \sin \left (d x + c\right ) - 4 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \] Input:
integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^2*tan(d*x+c)^2,x, algorithm="frica s")
Output:
-1/8*(6*a^2*sin(d*x + c) - 4*a^2 - (a^2*cos(d*x + c)^2 + 2*a^2*sin(d*x + c ) - 2*a^2)*log(sin(d*x + c) + 1) + (a^2*cos(d*x + c)^2 + 2*a^2*sin(d*x + c ) - 2*a^2)*log(-sin(d*x + c) + 1))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) - 2*d)
\[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=a^{2} \left (\int \tan ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)**3*(a+a*sin(d*x+c))**2*tan(d*x+c)**2,x)
Output:
a**2*(Integral(tan(c + d*x)**2*sec(c + d*x)**3, x) + Integral(2*sin(c + d* x)*tan(c + d*x)**2*sec(c + d*x)**3, x) + Integral(sin(c + d*x)**2*tan(c + d*x)**2*sec(c + d*x)**3, x))
Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.06 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac {2 \, {\left (3 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^2*tan(d*x+c)^2,x, algorithm="maxim a")
Output:
1/8*(a^2*log(sin(d*x + c) + 1) - a^2*log(sin(d*x + c) - 1) + 2*(3*a^2*sin( d*x + c) - 2*a^2)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d
Time = 0.19 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.88 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {1}{8} \, a^{2} {\left (\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{d} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} + \frac {2 \, {\left (3 \, \sin \left (d x + c\right ) - 2\right )}}{d {\left (\sin \left (d x + c\right ) - 1\right )}^{2}}\right )} \] Input:
integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^2*tan(d*x+c)^2,x, algorithm="giac" )
Output:
1/8*a^2*(log(abs(sin(d*x + c) + 1))/d - log(abs(sin(d*x + c) - 1))/d + 2*( 3*sin(d*x + c) - 2)/(d*(sin(d*x + c) - 1)^2))
Time = 33.62 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.81 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )} \] Input:
int((tan(c + d*x)^2*(a + a*sin(c + d*x))^2)/cos(c + d*x)^3,x)
Output:
(a^2*atanh(tan(c/2 + (d*x)/2)))/(2*d) - ((a^2*tan(c/2 + (d*x)/2)^3)/2 - 2* a^2*tan(c/2 + (d*x)/2)^2 + (a^2*tan(c/2 + (d*x)/2))/2)/(d*(6*tan(c/2 + (d* x)/2)^2 - 4*tan(c/2 + (d*x)/2) - 4*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x)/ 2)^4 + 1))
Time = 0.19 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.24 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {a^{2} \left (-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+3 \sin \left (d x +c \right )^{2}-1\right )}{8 d \left (\sin \left (d x +c \right )^{2}-2 \sin \left (d x +c \right )+1\right )} \] Input:
int(sec(d*x+c)^3*(a+a*sin(d*x+c))^2*tan(d*x+c)^2,x)
Output:
(a**2*( - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + 4*log(tan((c + d*x )/2) - 1)*sin(c + d*x) - 2*log(tan((c + d*x)/2) - 1) + 2*log(tan((c + d*x) /2) + 1)*sin(c + d*x)**2 - 4*log(tan((c + d*x)/2) + 1)*sin(c + d*x) + 2*lo g(tan((c + d*x)/2) + 1) + 3*sin(c + d*x)**2 - 1))/(8*d*(sin(c + d*x)**2 - 2*sin(c + d*x) + 1))